0

I need to prove, that sum of two autocovariance functions is an autocovariance function.

I take two random processes $X,Y$ for which $X_1(t),X_2(t)$ random variables are independent. Their autocovariance functions are respectfully $\Gamma_1$ and $\Gamma_2$. Then autocovariance of $X+Y$ is $\Gamma_1+\Gamma_2$. The idea was given during a lecture, but I don't know how to use it. Could anyone help me out? Thanks.

atomoutside
  • 493
  • 1
    I'm pretty sure you need independence of the whole processes $X$ and $Y$, and not just of $X(t)$ and $Y(t)$ for every $t$. If this is assumed, then this is shown using the bi-linearity of the covariance. – Stefan Hansen Jun 14 '13 at 05:49
  • Let's say that independence of $X$ and $Y$ is assumed. Then: $Cov(X,Y)=Cov(X_1+X_2,Y)=\ldots$ ? – atomoutside Jun 14 '13 at 05:58
  • 1
    What? You want to show that the autocovariance of $X+Y$ is of a certain form. Hence you should start with the definition of the autocovariance for $X+Y$: $\mathrm{Cov}(X(t)+Y(t),X(s)+Y(s))=??$ – Stefan Hansen Jun 14 '13 at 06:00
  • Stupid me... $$Cov(X(t)+Y(t),X(s)+Y(s))=Cov(X(t),X(s))+Cov(Y(t),Y(s))+Cov(X(t),Y(s))+Cov(Y(t),X(s))=Cov(X(t),X(s))+Cov(Y(t),Y(s)$$ due to fact, that $X$ and $Y$ are independent and thus uncorrelated? – atomoutside Jun 14 '13 at 06:18
  • Exactly :) You should post this as an answer and accept it, so that the question doesn't go unanswered. – Stefan Hansen Jun 14 '13 at 06:21

1 Answers1

1

Thanks to Stefan Hansen:

$$Cov(X(t)+Y(t),X(s)+Y(s))=Cov(X(t),X(s))+Cov(Y(t),Y(s))+Cov(X(t),Y(s))+Cov(Y(t)‌​,X(s))=Cov(X(t),X(s))+Cov(Y(t),Y(s)$$

due to fact, that $X$ and $Y$ are independent and thus uncorrelated.

Community
  • 1
atomoutside
  • 493